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WBJEE · Physics · Thermal Properties of Matter

\(A\) 10 W electric heater is used to heat a container filled with \(0.5 \mathrm{kg}\) of water. It is found that the temperature of water and the container rises by \(3^{\circ} \mathrm{K}\) in \(15 \mathrm{min}\). The container is then emptied, dried and filled with \(2 \mathrm{kg}\) of oil. The same heater now raises the temperature of container-oil system by \(2^{\circ} \mathrm{K}\) in \(20 \mathrm{min}\). Assuming that there is no heat loss in the process and the specific heat of water is \(4200 \mathrm{d} \mathrm{kg}^{-1} \mathrm{K}^{-1}\), the specific heat of oil in the same unit is equal to

  1. A \(1.50 \times 10^{3}\)
  2. B \(2.55 \times 10^{3}\)
  3. C \(3.00 \times 10^{3}\)
  4. D \(5.10 \times 10^{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2.55 \times 10^{3}\)

Step-by-step Solution

Detailed explanation

\(m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=\) Work done \(m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=P_{1} t_{1}\) where \(m_{1}=0.5 \mathrm{kg}\) Specific heat \(s_{1}=4200 \mathrm{J} / \mathrm{kg}-\mathrm{K}\)…